Question

Let $f\left(n\right)$ equals to the sum of the cubes of three consecutive natural numbers.
$f\left(n\right)$ leaves the remainder zero when divided by

• A. $11$
• B. $9$
• C. $99$
• D. none of these

Answer 1

Answer: (B)

$9$

Given that $f\left(n\right)={(n-1)}^3+n^3+{(n+1)}^3=3n^3+6n$
Put $n=1$, to obtain $f\left(1\right)={3.1}^3+6.1=9$
Therefore, $f\left(1\right)$ is divisible by $9$
Assume that for $n=k$, $f\left(k\right)=3k^3+6k$ is divisible by $9$
Now, $f(k+1)=3{(k+1)}^3+6(k+1)=3k^3+6k+9(k^2+k+1)=f\left(k\right)+9(k^2+k+1)$
Since, $f\left(k\right)$ is divisible by $9$
Therefore, $f(k+1)$ is divisible by $9$
And from the principle of mathematical induction $f\left(n\right)$ is divisible by $9$ for all $n\in N$
Hence, option B is correct.